Question

A material of dielectric constant 2 is inserted between the plates of a capacitor 3 micro F.
calculate the new value of the capacitance.

Answer 1

Given:

A material of dielectric constant 2 is inserted between the plates of a capacitor 3 micro F.

To find:

New value of capacitance ?

Calculation:

The general expression for capacitance (C) of a capacitor with a dielectric material (of Dielectric Constant K) placed inside it is given as :

$\boxed{ \bold{\therefore \: C_{(new)} = C_{(air)} K}}$

Putting the values in SI units:

$\implies \: C_{(new)} = (3 \times {10}^{ - 6} ) \times 2$

$\implies \: C_{(new)} = 6 \times {10}^{ - 6} \: F$

$\implies \: C_{(new)} = 6 \: \mu F$

So, the new capacitance of the capacitor is 6 micro F.